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$$\sum_{n=1}^{\infty} \frac{3}{\sqrt[3]{n^2+2}}$$

It seems clear to me that this series diverges because the dominant temr is $1/n^{2/3}$, a p-series with $p < 1$

However I need to prove divergence using something like the comparison test, integral test, or similar.

I can't work out a suitable comparison to make to prove divergence. Suggestions?

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  • $\begingroup$ The $p$-series test and the comparison test are pretty much the same (look at their proofs). Compare with $\left(\dfrac{3}{\root 3\of {n^2}}\right)_{n\in \mathbb N}$. $\endgroup$ – Git Gud May 21 '16 at 0:23
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Note that for all positive integers $n$, $$n^2 + 2 < n^2 + 4n + 4 = (n+2)^2.$$ Therefore, $$\frac{3}{\sqrt[3]{n^2 + 2}} > \frac{3}{\sqrt[3]{(n+2)^2}} > \frac{1}{(n+2)^{2/3}},$$ hence $$\sum_{n=1}^\infty \frac{3}{\sqrt[3]{n^2+2}} > \sum_{n=1}^\infty \frac{1}{(n+2)^{2/3}} = - 1^{2/3} - \frac{1}{2^{2/}} + \sum_{n=1}^\infty \frac{1}{n^{2/3}},$$ the last sum of which is a divergent $p$-series, hence the given sum also diverges.

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You have $${3\over \root{3}\of{n^2 + 2}}\sim {3\over{n^{2/3}}}.$$ Now apply the limit comparsion and $p$-series theorems.

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